Advertisements
Advertisements
प्रश्न
Empirical formula of a compound is CH2O. If its empirical formula is equal to its vapour density, calculate the molecular formula of the compound.
Advertisements
उत्तर
Empirical formula of the compound is CH2O.
Empirical formula mass = Atomic mass of C + Atomic mass of H + Atomic mass of O
= 12 + 2 x 1 + 16 = 30.
Now as empirical formula is equal to the vapour density then;
Molecular mass = 2 x vapour density
= 2 x 30 = 60
n = Molecular mass/Empirical formula mass
= 60 / 30 = 2
Molecular formula = n x empirical formula
= 2 x (CH2O) = C2H4O2
The molecular formula of the compound is C2H4O2
APPEARS IN
संबंधित प्रश्न
Potassium nitrate on strong heating decomposes as under :
2KNO3 → 2KNO2 + O2
Calculate : Weight of oxygen formed when 5.05g of potassium nitrate decomposes completely.
(K = 39, 0 = 16, N = 14)
What weight of sulphuric acid will be required to dissolve 3g of magnesium carbonate?
[Mg = 24, C =12, 0 = 16 ]
MgCO3 + H2SO4 → MgSO4 + H2O+ CO2
Washing soda has the formula Na2CO3.10H2O.What is the mass of anhydrous sodium carbonate left when all the water of crystallization is expelled by heating 57.2 g of washing soda?
When excess lead nitrate solution was added to a solution of sodium sulphate, 15.1g of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
Na2SO4 + Pb(NO3)2 → PbSO4 + 2NaNO3
(H = 1, C = 12, O = 16, Na = 23, S = 32, Pb = 207)
Chlorine, nitrogen, ammonia and sulphur dioxide gases are collected under the same conditions of temperature and pressure.
Copy the following table which gives the volumes of the gases collected, and the number of molecules (X) in 20L of nitrogen.You are to complete the table by giving the number of molecules in th e other gases, in terms of X.
| Gas | Volume(litres) | Number of molecules |
| Chlorine | 10 | |
| Nitrogen | 20 | X |
| Ammonia | 20 | |
| Sulphur dioxide | 5 |
A metal M, forms a volatile chloride containing 65.5% Chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride. [M = 56, Cl = 35.5]
The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below:
2KMno4 + 10FeSO4 + 8H2O → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O
If 15.8g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.
4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.
- Write the equation for the reaction.
- What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100).
- What is the volume of carbon dioxide liberated at STP?
- What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).
- How many moles of HCl are used in this reaction?
Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.
When heated, potassium permanganate decomposes according to the following equation :
\[\ce{2KMnO4 -> \underset{\text{solid residue}}{K2MnO4 + MnO2} + O2}\]
(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158. What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres)
